2. If \( x-i y=\sqrt{\frac{a-i b}{c-i d}} \) prove that \( \left(x^{2}+y^{2}\right)=\frac{a^{2}+b^{2}}{c^{2}+d^{2}} \)

To prove that \( x^2 + y^2 = \frac{a^2 + b^2}{c^2 + d^2} \) given \( x - iy = \sqrt{\frac{a - ib}{c - id}} \), let's start by expressing the right-hand side in terms of its modulus. 1. First, let \( z_1 = a - ib \) and \( z_2 = c - id \). The modulus of the quotient can be represented as: \[ \left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|} = \frac{\sqrt{a^2 + b^2}}{\sqrt{c^2 + d^2}} \] 2. Thus, the modulus of \( \sqrt{\frac{z_1}{z_2}} \) is: \[ \left| \sqrt{\frac{z_1}{z_2}} \right| = \sqrt{\left| \frac{z_1}{z_2} \right|} = \sqrt{\frac{|z_1|}{|z_2|}} = \sqrt{\frac{\sqrt{a^2 + b^2}}{\sqrt{c^2 + d^2}}} = \frac{\sqrt{a^2 + b^2}}{\sqrt{c^2 + d^2}} \] 3. The equation \( x - iy = \sqrt{\frac{a - ib}{c - id}} \) implies that: \[ |x - iy| = \left| \sqrt{\frac{a - ib}{c - id}} \right| \] 4. Therefore, we have: \[ \sqrt{x^2 + y^2} = \frac{\sqrt{a^2 + b^2}}{\sqrt{c^2 + d^2}} \] 5. Squaring both sides leads to: \[ x^2 + y^2 = \frac{a^2 + b^2}{c^2 + d^2} \] This completes the proof that \( x^2 + y^2 = \frac{a^2 + b^2}{c^2 + d^2} \).